Question

in figure ABCD. is a square of sides 14 cm semi circle are draw with each side of square as diaameter find the area of the shaded region​

Answer 1

Given:

ABCD is a square of side 14 cm.

Semi circles are drawn with each side of the square as diameter.

To find:

The area of the shaded region.

By observing the figure, we get four same parts.

Looking at individual four parts,

We can get two shaded part in it.

For one shaded part, we have to find the area of the whole part and minus one sector from it.

• To get EBO lined part,

Area of square ERFO - the area of sector OBF

$= (side)^{2} - \frac{\theta}{360}\times \pi r^{2}$

$= (7)^{2} - [\frac{90}{360}\times \frac{22}{7} \times (7)^{2}]$

$= 49 - [\frac{1}{4}\times \frac{22}{7} \times 49]$

$= 49- [\frac{1}{2}\times 11 \times 7]$

$= 49 - \frac{77}{2}$

= 49 - 38.5

= 10.5 cm sq.

Therefore,

For one lied portion area = 10.5 cm sq.

In one individual part, there are two so,

Area of two lined part = 10.5 * 2 = 21 cm sq.

Thus,

There are four such individual parts,

Area = 21 * 4 = 84 cm sq.

Hence,

• The area of the shaded region is 84 cm sq.

Answer 2

Answer:

Given :-

• ABCD, is a square whose side is 14 cm. Semicircle are draw with each side of square as a diameter.

To Find :-

• What is the area of the shaded region.

Solution :-

Let, the four shaded region be first, second, third and fourth are the centers of the semicircle will be P, Q, R and S.

Given :

• Semicircle = 14 cm.
• Radius of the semicircle = $\sf\dfrac{14}{2}$ cm = 7 cm

Now, according to the question,

✪ Area of first region + Area of third region = Area of the square - Area of the semicircle ✪

⇒ $\sf {(14)}^{2} - 2 \times \dfrac{1}{2} \times {\pi}{r}^{2}$

⇒ $\sf 14 \times 14 - 2 \times \dfrac{1}{2} \times \dfrac{22}{7} \times {(7)}^{2}$

⇒ $\sf 196 - 2 \times \dfrac{1}{2} \times \dfrac{22}{7} \times 49$

⇒ $\sf 196 - 2 \times \dfrac{1}{2} \times 154$

⇒ $\sf 196 - 154$

➠ $\sf 42\: {cm}^{2}$

Again,

★ Area of second region + Area of fourth region = Area of the square - Area of the semicircle ★

↦ $\sf {(14)}^{2} - 2 \times \dfrac{1}{2} \times {\pi}{r}^{2}$

↦ $\sf 14 \times 14 - 2 \times \dfrac{1}{2} \times \dfrac{22}{7} \times {(7)}^{2}$

↦ $\sf 196 - 2 \times \dfrac{1}{2} \times \dfrac{22}{7} \times 49$

↦ $\sf 196 - 154$

➦ $\sf 42\: {cm}^{2}$

Now, we have to find the area of the shaded region,

➤ Area of the shaded region = Area of first region + Area of third region + Area of the second region + Area of the fourth region.

↪ 42 cm² + 42 cm²

➲ 84 cm²

$\therefore$ The area of the shaded region is 84 cm² .